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Chapter 3: Problem 10

In Exercises \(7-12,\) find a linearization at a suitably chosen integer near \(a\) at which the given function and its derivative are easy to evaluate. $$ f(x)=1+x, \quad a=8.1 $$

Short Answer

Expert verified
The linearization of \(f(x) = 1 + x\) at \(a = 8\) is \(L(x) = x + 1\).

Step by step solution

01

Choose the Suitable Integer

First, determine the integer near the given point \(a = 8.1\) that will make computations easier. The closest integer to 8.1 is 8. Therefore, choose \(a = 8\) for linearization.
02

Find the Function and its Derivative

The function given is \(f(x) = 1 + x\). Calculate its derivative, \(f'(x)\). Since \(f(x) = 1 + x\), the derivative is simply \(f'(x) = 1\).
03

Evaluate at the Integer Chosen

Evaluate both the function and its derivative at \(a = 8\). Here, \(f(8) = 1 + 8 = 9\) and \(f'(8) = 1\).
04

Write the Linearization Formula

The linearization \(L(x)\) of a function at a chosen point \(a\) is given by the formula: \[L(x) = f(a) + f'(a)(x - a)\].Substitute the values found: \(f(a) = 9\), \(f'(a) = 1\), and \(a = 8\).
05

Substitute Values into the Linearization Formula

Use the values calculated in Step 3 in the linearization formula: \[L(x) = 9 + 1(x - 8)\].This simplifies to \(L(x) = 9 + x - 8\), or equivalently, \(L(x) = x + 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Calculus
Calculus is a branch of mathematics that focuses on understanding how things change. It encompasses two primary operations: differentiation and integration. Differentiation deals with finding things like rates of change, while integration helps in finding accumulated quantities.

In simple terms, calculus allows us to analyze and model the behavior of curves and various physical, biological, and economic processes. It's like a toolkit for tackling problems that involve change. The power of calculus lies in its ability to handle infinite detail without confusion. For our exercise, we focus on linearization, an application of differentiation.
Understanding Function Derivative
The derivative of a function is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. For our function, \( f(x) = 1 + x \), the derivative is \( f'(x) \).

The process of finding a derivative is known as differentiation. For our function, the derivative is simply \( f'(x) = 1 \) because the change in \( x \) results in an equal change in \( f(x) \).
  • The derivative helps us understand the function's behavior.
  • It tells if the function is increasing or decreasing.
  • For linear functions, like \( f(x) = 1 + x \), the derivative is constant, meaning the growth rate is steady.
Understanding how to calculate a derivative is crucial, as it forms the basis for linear approximation.
The Concept of Linear Approximation
Linear approximation is a method of estimating the value of a function near a point. It uses the tangent line to approximate the function's value. This is particularly useful when a simpler function can replace a more complex one around a point.

To perform linear approximation, we use the linearization formula:\[L(x) = f(a) + f'(a)(x - a)\]
This formula tells us the value of the function \( f(x) \) near \( a \) by using the function's value and slope at \( a \).
  • This approach helps in simplifying calculations.
  • It provides a close approximation using the tangent line.
  • Ideal for getting quick estimates in real-world problems.
In our exercise, the linearization \( L(x) = x + 1 \) provides an approximate value of \( f(x) \) near \( a = 8 \), effectively turning the curve into a straight line for ease of use.
Integer Approximation and Its Role
Integer approximation involves selecting integers to simplify mathematical calculations. This concept is especially helpful in linearization because integers are easier to work with.

In our exercise, we chose \( a = 8 \) as it is the closest integer to 8.1. This makes it easier to calculate the function and its derivative, allowing for a simpler approximation. By using integer approximation:
  • We simplify calculations.
  • The approximations are easier and faster.
  • Minimizes error in utterly complex problems.
Integer approximation is a crucial step in making linearization practical, especially when exact solutions are cumbersome or impossible to obtain with precision.

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Most popular questions from this chapter

Changing dimensions in a rectangle The length \(l\) of a rectangle is decreasing at the rate of 2 \(\mathrm{cm} / \mathrm{sec}\) while the width \(w\) is increasing at the rate of 2 \(\mathrm{cm} / \mathrm{sec} .\) When \(l=12 \mathrm{cm}\) and \(w=5 \mathrm{cm},\) find the rates of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing, and which are increasing?

In Exercises \(35-40,\) write a differential formula that estimates the given change in volume or surface area. $$ \begin{array}{l}{\text { The change in the volume } V=\pi r^{2} h \text { of a right circular cylinder }} \\ {\text { when the radius changes from } r_{0} \text { to } r_{0}+d r \text { and the height does }} \\ {\text { not change }}\end{array} $$

Motion in the plane The coordinates of a particle in the metric \(x y\)-plane are differentiable functions of time \(t\) with \(d x / d t=\) \(-1 \mathrm{m} / \mathrm{sec}\) and \(d y / d t=-5 \mathrm{m} / \mathrm{sec} .\) How fast is the particle's distance from the origin changing as it passes through the point \((5,12) ?\)

A growing raindrop Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate.

A draining hemispherical reservoir Water is flowing at the rate of 6 \(\mathrm{m}^{3} / \mathrm{min}\) from a reservoir shaped like a hemispherical bowl of radius \(13 \mathrm{m},\) shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radius \(R\) is \(V=(\pi / 3) y^{2}(3 R-y)\) when the water is \(y\) meters deep. a. At what rate is the water level changing when the water is 8 m deep? b. What is the radius \(r\) of the water's surface when the water is \(\quad y\) m deep? c. At what rate is the radius \(r\) changing when the water is 8 \(\mathrm{m}\) deep?
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